Question

A tensile load of 40KN is acting on a rod of diameter 40mm and length 4mm.A bone of diameter 20mm is made centrally on the rod.To what lenght the rod should bored so that total extension will increase 30 percentage under the same tensile load.Take E=2×10^5N/mm​

Answer 1

Step-by-step explanation:

I didn't no the answer plz ask to anyone

Answer 2

Given :

Tensile load,              $\[P = 40\,kN = 40000\,N\]$

Dia of the rod,           $\[D = 40\,mm\]$

∴ Area of the rod,      $\[A = \frac{\pi }{4}\left( {{{40}^2}} \right) = 400\pi \,m{m^2}\]$

Length of the rod,      $\[L = 4m = 4000mm\]$

Dia. of the bore,         $\[d = 20mm\]$  

∴ Area of the bore,     $\[a = \frac{\pi }{4}\left( {{{20}^2}} \right) = 100\pi \,m{m^2}\]$

To Find :

We have to find to what length, the rod should be bored so that total extension will increase to 30 percent.

Solution:

Let the rod be bored to a length of x meter. Refer to the attached figure.

∴ Length of the unbored portion   $\[ = \left( {4 - x} \right)\,m = (4 - x) \times 1000\,mm\]$

Total extension after bore = 1·3 × Extension before bore

Extension, before bore is made is given by,

                                          $\[\begin{array}{c}\Delta L = \frac{{P \times L}}{{AE}}\\\\ = \frac{{40000 \times 4000}}{{400\pi \times 2 \times {{10}^5}}}\\\\ = {\raise0.7ex\hbox{ 2 } \!\mathord{\left/ {\vphantom {2 \pi }}\right.\kern-\nulldelimiterspace}\!\lower0.7ex\hbox{ \pi }}\,mm\end{array}\]$

Extension, after bore is made = 1·3 × Extension before bore

                                                 $\[ = 1.3 \times \frac{2}{\pi } = \frac{{2.6}}{\pi }\,mm\]$

The extension after the bore is made, is also obtained from the sum of the extensions of the bored and unbored length.

Extension of the unbored portion,

                                                        $\[ = \frac{P}{{AE}} \times Length\,of\,unbored\,portion\]$

                                                        $\[\begin{array}{l} = \frac{{40000}}{{400\pi \times 2 \times {{10}^5}}} \times \left( {4 - x} \right) \times 1000\\\\ = \frac{{4 - x}}{{2\pi }}\,mm\end{array}\]$  

Extension of the bored portion,    

                                                        $\[ = \frac{P}{{\left( {A - a} \right)}} \times length\,of\,bored\,portion\]$

                                                        $\[\begin{array}{l} = \frac{{40000}}{{300\pi \times 2 \times {{10}^5}}} \times 1000x\\\\ = \frac{{4x}}{{6\pi }}\,mm\end{array}\]$

∴ Total extension after the bore is made                                      

                                                        $\[ = \frac{{\left( {4 - x} \right)}}{{2\pi }} + \frac{{4x}}{{6\pi }}\]$    

Equating the equations ⇒

                                          $\[\begin{array}{c}\frac{{2 \cdot 6}}{\pi } = \frac{{\left( {4 - x} \right)}}{{2\pi }} + \frac{{4x}}{{6\pi }}\\\\2 \cdot 6 = \frac{{\left( {4 - x} \right)}}{{2\pi }} + \frac{{4x}}{{6\pi }}\\\\2 \cdot 6 \times 6 = 3 \times \left( {4 - x} \right) + 4x\\\\15 \cdot 6 = 12 - 3x + 4x\\\\15 \cdot 6 - 12 = x\\\end{array}\]$

                                                      $\[x = 3.6\]$  

∴ The rod should be bored up to a length of 3.6 m.